OC and SC tests to build transformer equivalent circuit

The no-load (open-circuit) test and short-circuit test find the equivalent-circuit parameters and losses of the 10 kVA, 6600/220 V transformer without loading it. Rated currents: HV $= 10000/6600 = 1.515$ A; LV $= 10000/220 = 45.45$ A.

No-load (OC) test

  0-220 V  +--(A)--[W]--+------+     +-----o
  variac   |            |  LV  | ||  |  HV  open
  50 Hz   (V)           | 220V | ||  | 6600 V
           +------------+------+     +-----o

• Done on the LV (220 V) side, HV open. The LV side is used because 220 V is safe and easy to supply and measure; the HV terminals still have 6600 V on them and must be guarded.
• Apply rated voltage, 220 V, at rated frequency. Read $V_0$, $I_0$, $P_0$.
• $I_0$ is small (about 2-5% of 45.45 A, i.e. 1-2 A), so copper loss is negligible and $P_0$ = iron loss.

Calculations:

$$\cos\phi_0 = \frac{P_0}{V_0I_0}, \quad I_w = I_0\cos\phi_0, \quad I_\mu = I_0\sin\phi_0, \quad R_0 = \frac{V_0}{I_w}, \quad X_0 = \frac{V_0}{I_\mu}$$

These are on the LV side.

Short-circuit test

  low V    +--(A)--[W]--+------+     +-----+
  variac   |            |  HV  | ||  |  LV | thick
  50 Hz   (V)           | 6600V| ||  | 220V| short
           +------------+------+     +-----+

• Done on the HV (6600 V) side, LV shorted by a thick link. The HV side is used because rated current there is only 1.515 A and the needed voltage (5-10% of 6600 V, i.e. about 330-660 V) is easy to measure; on the LV side we would need 45 A at only 11-22 V.
• Raise voltage slowly from zero until rated current 1.515 A flows. Read $V_{sc}$, $I_{sc}$, $P_{sc}$.
• Flux is very low, so iron loss is negligible and $P_{sc}$ = full-load copper loss.

Calculations (HV side, which is the primary here):

$$Z_{01} = \frac{V_{sc}}{I_{sc}}, \quad R_{01} = \frac{P_{sc}}{I_{sc}^2}, \quad X_{01} = \sqrt{Z_{01}^2 - R_{01}^2}$$

Developing the equivalent circuit referred to primary (6600 V)

1. $R_{01}$ and $X_{01}$ come directly from the SC test because it was on the HV (primary) side.
2. $R_0$ and $X_0$ from the OC test are on the LV side, so refer them to the primary by multiplying by

$$\left(\frac{6600}{220}\right)^2 = 30^2 = 900$$

i.e. $R_0' = 900R_0$, $X_0' = 900X_0$. 3. Draw the approximate equivalent circuit: shunt branch ($R_0' \parallel X_0'$) across the 6600 V input, followed by $R_{01} + jX_{01}$ in series, and load referred to primary $Z_L' = 900Z_L$.

 o---+---------+---[ R01 ]---[ jX01 ]---o
     |         |                        +
  [ R0' ]   [ jX0' ]               V2' = 30 V2
     |         |                        -
 o---+---------+------------------------o

Example: if the OC test gives 220 V, 1.2 A, 90 W, then $\cos\phi_0 = 0.341$, $I_w = 0.409$ A, $R_0 = 538\ \Omega$ (LV), so $R_0' = 484\ \text{k}\Omega$ on the 6600 V side.

From the same results: iron loss and full-load copper loss give efficiency at any load, and $R_{01}$, $X_{01}$ give voltage regulation.